Compute $\tan (-3645^\circ)$.
Rotating $360^\circ$ is the same as doing nothing, so rotating $3645^\circ$ is the same as rotating $3645^\circ - 10\cdot 360^\circ = 45^\circ$.  Therefore, $\tan(-3645^\circ) = \tan (-45^\circ)$.

Rotating $45^\circ$ clockwise is the same as rotating $360^\circ - 45^\circ = 315^\circ$ counterclockwise, so $\tan(-45^\circ) = \tan (360^\circ - 45^\circ) = \tan 315^\circ$.

Let $P$ be the point on the unit circle that is $315^\circ$ counterclockwise from $(1,0)$, and let $D$ be the foot of the altitude from $P$ to the $x$-axis, as shown below.

[asy]

pair A,C,P,O,D;

draw((0,-1.2)--(0,1.2),p=black+1.2bp,Arrows(0.15cm));

draw((-1.2,0)--(1.2,0),p=black+1.2bp,Arrows(0.15cm));

A = (1,0);

O= (0,0);

label("$x$",(1.2,0),SE);

label("$y$",(0,1.2),NE);

P = rotate(315)*A;

D = foot(P,A,-A);

draw(O--P--D);

draw(rightanglemark(O,D,P,2));

draw(Circle(O,1));

label("$O$",O,NW);

label("$P$",P,SE);

//label("$A$",A,SE);

label("$D$",D,N);

[/asy]

Triangle $POD$ is a 45-45-90 triangle, so $DO = OP = \frac{\sqrt{2}}{2}$.  Therefore, the coordinates of $P$ are $\left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)$, so $\tan(-3645^\circ) = \tan (-45^\circ) = \tan 315^\circ = \frac{\sin 315^\circ}{\cos 315^\circ} = \frac{-\sqrt{2}/2}{\sqrt{2}/2} = \boxed{-1}$.